Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reached in their paths the samemaximum height. Ball A covered the total horizontal range D; ball B covered the total horizontal range 1.5D and ball C covered total horizontal range 2D when they hit the ground. Which of these three balls spent the greatest time in flight? (section 4.3) They all have the same time of flight ОА OB oc Band A and B

Answer:

So car will go 50 km in 15 minutes

Explanation:

We have given speed of the car = 200 km/hour

Time t = 15 minutes

We know that 1 hour = 60 minute

So 15 minute = [tex]\frac{15}{60}=0.25hour[/tex]

We have to find the distance

We know that distance = speed ×time = 200×0.25=50 km

So car will go 50 km in 15 minutes

So option (d) will be the correct option

Answer:

The longest wavelength equals [tex]0.4\times 10^{-6}m[/tex]

Explanation:

According to Einstein's photoelectric equation we have

[tex]E_{incident}\geq \phi [/tex]

where

[tex]E_{incident}[/tex] is the energy of the incident light

[tex]\phi [/tex] is the work function of the metal

The incident energy of the light with wavelength [tex]\lambda [/tex] is given by

[tex]E_{incident}=h\cdot \frac{c}{\lambda}[/tex]

Thus the photoelectric equation reduces to

[tex]h\cdot \frac{c}{\lambda}\geq \phi\\\\h\cdot c\geq \lambda \times \phi\\\\\therefore \lambda\leq \frac{h\cdot c}{\phi}[/tex]

Thus applying values we get

[tex]\lambda\leq \frac{6.62\times 10^{-34}\times 3\times 10^{8}}{3.10\times 1.602\times 10^{-19}}\\\\\therefore \lambda\leq 0.4\times 10^{-6}m[/tex]

Hence The longest wavelength equals [tex]0.4\times 10^{-6}m[/tex]

Answer:

Magnitude of electric field at mid way is [tex]5.07\times 10^{7} N/C[/tex]

Solution:

As per the question:

Q = [tex]- 5.6\mu m = - 5.6\times 10^{- 6} C[/tex]

Q' = [tex]5.8\mu m = 5.8\times 10^{- 6} C[/tex]

Separation distance, d = 9.0 cm = 0.09 m

The distance between charges at mid-way, O is [tex]\farc{d}{2} = 0.045 m[/tex]

Now, the electric at point O due to charge Q is:

[tex]E = \frac{1}{4\pi \epsilon_{o}}.\frac{Q}{(\frac{d}{2})^{2}}[/tex]

[tex]E = \frac{1}{4\pi \epsilon_{o}}.\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}[/tex]

[tex]E = (9\times 10^{9})\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}[/tex]

[tex]E =  (9\times 10^{9})\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}[/tex]

[tex]E = - 2.49\times 10^{7} N/C[/tex]

Here, negative sign is indicative of the direction of electric field which is towards the point O

Now, the electric at point O due to charge Q' is:

[tex]E' = \frac{1}{4\pi \epsilon_{o}}.\frac{Q}{(\frac{d}{2})^{2}}[/tex]

[tex]E' =  (9\times 10^{9})\frac{5.8\times 10^{- 6}}{(0.045^{2}}[/tex]

[tex]E' = 2.58\times 10^{7} N/C[/tex]

Refer to Fig 1.

Since, both the fields are in the same direction:

[tex]E_{net} = E + E' = 2.49\times 10^{7} + 2.58\times 10^{7} = 5.07\times 10^{7} N/C[/tex]

The magnitude of the electric field at a point midway between two charges can be calculated using the formula E = k * (Q1-Q2) / r^2. Plugging in the values -5.6μC, +5.8μC, and 9.0cm, the magnitude of the electric field is -2.03 x 10^4 N/C.

The magnitude of the electric field at a point midway between two charges can be calculated using the formula:

E = k * (Q1-Q2) / r^2

where E is the electric field, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.

In this case, the charges are -5.6μC and +5.8μC, and the distance is 9.0cm (0.09m).

Plugging these values into the formula:

E = (9 x 10^9 Nm^2/C^2) * ((-5.6μC) - (+5.8μC)) / (0.09m)^2

Simplifying the equation:

E = -2.03 x 10^4 N/C

Answer:

Explanation:

The spacetime interval [tex]\Delta s^2[/tex] is given by

[tex]\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2[/tex]

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

[tex]\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2[/tex].

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

[tex]\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2[/tex]

[tex]\Delta \vec{x}^2 = 5,625 m^2[/tex]

[tex]\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2[/tex]

[tex]\Delta (c t) ^ 2 = (899,377,374 \ m)^2[/tex]

[tex]\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2[/tex]

so

[tex]\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2[/tex]

[tex]\Delta s ^2 = 8.0888 \ 10^{17} m^2[/tex]

[tex]\Delta \vec{x}^2 = (5 \ 10 \ m)^2[/tex]

[tex]\Delta \vec{x}^2 = 2,500 m^2[/tex]

[tex]\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2[/tex]

[tex]\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2[/tex]

[tex]\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2[/tex]

so

[tex]\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2[/tex]

[tex]\Delta s ^2 = 3.0234 \ 10^{16} m^2[/tex]

[tex]\Delta \vec{x}^2 = (5 \ 10 \ m)^2[/tex]

[tex]\Delta \vec{x}^2 = 2,500 m^2[/tex]

[tex]\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2[/tex]

[tex]\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2[/tex]

[tex]\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2[/tex]

so

[tex]\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2[/tex]

[tex]\Delta s ^2 = 3.0234 \ 10^{20} m^2[/tex]

Answer:

a) v = 15.6 m/s

b) 0.65 s are needed to reach a height of 12.3 m

Explanation:

The equations that describe the height and velocity of the stone are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where

y = height of the stone at time t

y0 = initial height

v0 = initial speed

t = time

g = acceleration due to gravity

b) First, let´s find the time at which the stone reaches a height of 12.3 m:

y = y0 + v0 · t + 1/2 · g · t²

12.3 m = 0 m + 22.0 m/s · t + 1/2 · (-9.8 m/s²) · t²    (y0 = 0 placing the center of the frame of reference at the point at which the stone is thrown.)

-4.9 m/s² · t² + 22.0 m/s · t - 12.3 m = 0

t = 0.65 s (when the stone goes upward) and t = 3.84 s ( when the stone returns downward) .

So, 0.65 s are needed to reach a height of 12.3 m

a) The velocity at that time will be:

v = v0 + g · t

v = 22.0 m/s - 9.8 m/s² · 0.65 s = 15.6 m/s

The stone moves at approximately 15.62 m/s when it reaches a height of 12.3 meters. The time required to reach this height is approximately 0.652 seconds.

Initial velocity (u) = 22.0 m/s

Acceleration (a) = -9.8 m/s² (due to gravity)

Height (h) = 12.3 m

We can use the equation:

v² = u² + 2a(s), where v is the final velocity, u is the initial velocity, a is acceleration, and s is the displacement.

Substituting the given values:

v² = (22.0 m/s)² + 2(-9.8 m/s²)(12.3 m)

v² = 484 - 240.12

v² = 243.88

v = √243.88 ≈ 15.62 m/s

Thus, the stone is moving at approximately 15.62 m/s when it reaches a height of 12.3 m.

We can use the equation:

s = ut + 1/2 at², where s is the displacement, u is the initial velocity, t is time, and a is acceleration.

12.3 m = (22.0 m/s)t + 1/2(-9.8 m/s²)t²

12.3 = 22.0t - 4.9t²

Rearrange to form a quadratic equation:

-4.9t² + 22.0t - 12.3 = 0

Using the quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

t = [ 22.0 ± √(484 - 4(-4.9)(-12.3)) ] / 2(-4.9)

t = ( 22.0 ± √(484 - 240.12) ) / 9.8

t = ( 22.0 ± √243.88 ) / 9.8

t = 22.0 ± 15.62 / 9.8

We get two possible values for t:

t₁ = (22.0 - 15.62) / 9.8 ≈ 0.652 s

t₂ = (22.0 + 15.62) / 9.8 ≈ 3.83 s

The correct time to reach 12.3 m as the stone ascends is approximately 0.652 seconds.

Answer:

Explanation:

The pressure of the gas can be found out as follows

Gas law formula is as follows

PV = nRT

P = nRT / V

= 32.4 X 8.31 X ( 273 + 459.38 ) / 51 X 10⁻⁶

3866.45 X 10⁶ Pa.

The first change is isobaric therefore

V₁ / T₁ = V₂ / T₂

V₂ = V₁ X T₂/ T₁

= 51 X 10⁻⁶ X ( 855.6 +273) / (459.38 +273)

= 78 X 10⁻⁶

78 cm³

After the first operation , the pressure of the gas remains at

= 3866.45 X 10⁶ Pa.

Now volume of the gas changes from 78 cm³ to 25.3 cm³ isothermally so

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

= 3866.45 X 10⁶ X 78 / 25.3

= 11920 X 10⁶ . Pa

Answer:

a) 2.41 km

b) 38.8°

Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

When he traveled south he moved to (-3, 0).

When he moved east he moved to (-3, x)

The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

dy^2 = d^2 - dx^2

[tex]dy = \sqrt{d^2 - dx^2}[/tex]

[tex]dy = \sqrt{3.85^2 - 3^2}  = 2.41 km[/tex]

The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

Answer:

Explanation:

A) we know that volume is given as V

[tex]V  =\frac{\pi}{4} D^2 h[/tex]

where D = 1.5 in , h = 2.0 in

so [tex]V = \frac{\pi}{4} 1.5^2\times 2 = 3.53in^3[/tex]

[tex]Area =\frac{\pi}{4} D^2 = \frac{\pi}{4} \times 1.5^2 = 1.76 in^4[/tex]

yield strenth is given as[tex] \sigma_y = \frac{force}{area} = \frac{140,000}{1.76}[/tex]

[tex]\sigma_y = 79.224 ksi[/tex]

b)

elastic strain[tex] \epsilon = \frac{\sima_y}{E} = \frac{79.224}{30\times 10^3} = 0.00264[/tex]

strain offsets  = 0.00264 + 0.002 = 0.00464     [where 0.002 is offset given]

[tex]\frac{\delta}{h} = 0.00464[/tex]

[tex]\frac{h_i -h_o}{h_o} = 0.00464[/tex]

[tex]h_i = 2\times(1-0.00464) = 1.99 inch[/tex]

area [tex]A = \frac{volume}{height} = \frac{3.534}{1.9907} = 1.775 in^2[/tex]

True strain[tex] \sigma = \frac{force}{area} = \frac{140,000}{1.775 in^2} = 78,862 psi[/tex]

At P= 260,000 lb ,[tex] A = \frac{3.534}{1.6} = 2.209 inc^2[/tex]

true stress [tex]\sigma  = \frac{260,000}{2.209} = 117,714 psi[/tex]

true strain [tex]\epsilon = ln\frac{2}{1.6} = 0.223[/tex]

flow curve is given as \sigma = k\epsilon^n

[tex]\sigma_1 = 78,862 psi[/tex]

[tex]\epsilon_1 = 0.00464[/tex]

[tex]\sigma_2 = 117,714 psi[/tex]  

[tex]\epsilon_2 = 0.223[/tex]

so flow curve is

[tex]78,868 = K 0.00464^n[/tex] .........1

[tex]117,714 = K 0.223^n[/tex]   .........2

Solving 1 and 2

we get

n = 0.103

and K =137,389 psi

Strength coffecient = K = 137.389ksi

strain hardening exponent = n = 0.103

Answer:

a) 138.6 m/s

b) 762.3 m

c) 122.3 m/s

d) 24.47

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

[tex]v=u+at\\\Rightarrow v=0+12.6\times 11\\\Rightarrow v=138.6 \ m/s[/tex]

Velocity at the end of its upward acceleration is 138.6 m/s

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 12.6\times 11^2\\\Rightarrow s=762.3\ m[/tex]

Maximum height the rocket reaches is 762.3 m

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 762.3-0^2}\\\Rightarrow v=122.3\ m/s[/tex]

The velocity with which the rocket crashes to the Earth is 122.3 m/s

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 762.3=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{762.3\times 2}{9.81}}\\\Rightarrow t=12.47\ s[/tex]

Total time from launch to crash is 12.47+11 = 24.47 seconds

Final answer:

The rocket's velocity at the end of its upward acceleration is 138.6 m/s. The maximum height it reaches is 1353.2 m. The rocket crashes to Earth with a velocity of -1.0 m/s.

Explanation:

(a)  To find the velocity at the end of the rocket's upward acceleration, we can use the formula:

v = u + at

Where:
u = initial velocity = 0 m/s (since the rocket starts from rest)
a = acceleration = 12.6 m/s2
t = time = 11.0 s

Substituting the values, we get:

v = 0 + 12.6 * 11.0 = 138.6 m/s

Therefore, the velocity at the end of its upward acceleration is 138.6 m/s.

(b)  To find the maximum height the rocket reaches, we can use the second equation of motion:

s = ut + 0.5at2

Where:
s = distance
u = initial velocity
a = acceleration
t = time

In this case, we need to consider the time taken for both the upward acceleration and free fall.

For the upward acceleration:

u = 0 m/s (since the rocket starts from rest)
a = 12.6 m/s2
t = 11.0 s

Substituting the values, we get:

s1 = 0 * 11.0 + 0.5 * 12.6 * (11.0)2 = 388.5 m

For the free fall:

u = 138.6 m/s (velocity at the end of the upward acceleration)
a = -9.8 m/s2 (acceleration due to gravity)
t = ?

To find the time for free fall, we can use the equation:

u = at

Substituting the values, we get:

138.6 = -9.8t

Solving for t, we get:

t = -14.1 s

However, time cannot be negative in this case. So, we take the absolute value of t:

t = 14.1 s

Substituting the values in the equation for free fall distance, we get:

s2 = 138.6 * 14.1 + 0.5 * (-9.8) * (14.1)2 = 964.7 m

The maximum height reached by the rocket is s1 + s2 = 388.5 m + 964.7 m = 1353.2 m.

(c)  To find the velocity at which the rocket crashes to Earth, we again consider the free fall phase. Using the equation:

v = u + at

Where:
u = 138.6 m/s (velocity at the end of the upward acceleration)
a = -9.8 m/s2 (acceleration due to gravity)
t = 14.1 s

Substituting the values, we get:

v = 138.6 - 9.8 * 14.1 = -1.0 m/s

The velocity at which the rocket crashes to Earth is -1.0 m/s. The negative sign indicates that the velocity is directed downward.

(d)  The total time from launch to crash is the sum of the time for upward acceleration (11.0 s) and the absolute value of the time for free fall (14.1 s). Therefore, the total time is 11.0 s + 14.1 s = 25.1 s.

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

[tex]X(van)=5.65t+154[/tex]

[tex]X(driver)=34.4t+\frac{(-2)t^{2} }{2}[/tex]

or by rearanging the drivers equation.

[tex]X(driver)=34.4t+t^{2}[/tex]

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

[tex]X(van)=X(driver)[/tex]

[tex]5.65t+154=34.4t-t^{2}[/tex]

[tex]0=t^{2} -(34.4-5.65)t+154[/tex][tex]0=t^{2} -28.75t+154[/tex]

To solve this equation we use the following formulas

[tex]t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}[/tex]

[tex]t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}[/tex]

Where a=1; b=-28.75; c=154

So we get:

[tex]t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s[/tex][tex]t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s[/tex]

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

[tex]V(driver)=V_{0} +at[/tex]}

[tex]V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}[/tex][tex]V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}[/tex]

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer  A).

Best of luck

The electron can make angles of 118° and 62° with the magnetic field.

An electron moving at a speed of 4.00 × 10³ m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40 × 10-16 N. What angle does the velocity of the electron make with the magnetic field?

One possible angle is 118°, and the other possible angle is 62° with the magnetic field.

The correct answer is [tex]\(\boxed{89.9999999999999\°}\).[/tex]

To determine the angle between the electron's velocity and the magnetic field, we can use the formula for the magnitude of the magnetic force on a moving charge, which is given by the Lorentz force law:

[tex]\[ F = qvB \sin(\theta) \][/tex]

where:

- [tex]\( F \)[/tex] is the magnitude of the magnetic force,

- [tex]\( q \)[/tex] is the charge of the electron,

- [tex]\( v \)[/tex] is the speed of the electron,

- [tex]\( B \)[/tex] is the magnitude of the magnetic field, and

- [tex]\( \theta \)[/tex] is the angle between the electron's velocity and the magnetic field.

The charge of an electron is [tex]\( 1.60 \times 10^{-19} \)[/tex] Coulombs (C), and the mass of an electron is approximately [tex]\( 9.11 \times 10^{-31} \)[/tex] kilograms (kg). Using Newton's second law, [tex]\( F = ma \)[/tex], where [tex]\( m \)[/tex] is the mass of the electron and [tex]\( a \)[/tex] is its acceleration, we can equate the magnetic force to the mass times acceleration:

[tex]\[ qvB \sin(\theta) = ma \][/tex]

Given that the electron experiences only a magnetic force, we can solve for [tex]\( \sin(\theta) \)[/tex]:

[tex]\[ \sin(\theta) = \frac{ma}{qvB} \][/tex]

Plugging in the given values:

[tex]\[ \sin(\theta) = \frac{(9.11 \times 10^{-31} \text{ kg})(2.30 \times 10^{14} \text{ m/s}^2)}{(1.60 \times 10^{-19} \text{ C})(7.69 \times 10^{6} \text{ m/s})(9.21 \times 10^{-4} \text{ T})} \][/tex]

[tex]\[ \sin(\theta) = \frac{(9.11 \times 2.30)}{(1.60 \times 7.69 \times 9.21)} \times 10^{-31 + 14 - (-19) - 6 - (-4)} \][/tex]

[tex]\[ \sin(\theta) = \frac{(20.953)}{(119.9424)} \times 10^{-31 + 14 + 19 - 6 + 4} \][/tex]

[tex]\[ \sin(\theta) = 0.1747 \times 10^{-31 + 14 + 19 - 6 + 4} \][/tex]

[tex]\[ \sin(\theta) = 0.1747 \times 10^{-10} \][/tex]

[tex]\[ \sin(\theta) = 1.747 \times 10^{-11} \][/tex]

Now, we can find the angle [tex]\( \theta \)[/tex] by taking the inverse sine (arcsin) of [tex]\( \sin(\theta) \)[/tex]:

[tex]\[ \theta = \arcsin(1.747 \times 10^{-11}) \][/tex]

Since the value of [tex]\( \sin(\theta) \)[/tex] is extremely small, the angle [tex]\( \theta \)[/tex] will be very close to 0 degrees. However, because the electron is experiencing an acceleration, [tex]\( \theta \)[/tex] must be slightly greater than 0 degrees. Using a calculator, we find:

[tex]\[ \theta \approx \boxed{89.9999999999999\°} \][/tex]

This result indicates that the electron's velocity is nearly parallel to the magnetic field, with an angle that is almost 90 degrees but infinitesimally less.

Answer:

Δf=73Hz

Explanation:

From the question we know that:

C = 1482 m/s

Vs = 0 m/s

Vr = -4.95 m/s (it's negative because it is in the opposite direction to the waves)

f0 = 22000 Hz

Applying the formula for the doppler effect:

[tex]f=(\frac{C-Vr}{C-Vs} )*fo[/tex]

f = 22073 Hz.   So the difference is only 73Hz

Answer:

Explanation:

At the topmost position,  the car does not have zero velocity but it has velocity of v so that

v² /r = g or centripetal acceleration should be equal to g ( 9.8 )

Considering that,  the car must fall from a height of 2r + h where

mgh = 1/2 mv²

= 1/2 m gr

So h = r/2

Hence the ball must fall from a height of

2r + r /2

= 2.5 r . So that it can provide velocity of v  at the top where

v² / r = g .

Explanation:

Given that,

Charge 1, [tex]q_1=-5.45\times 10^{-6}\ C[/tex]

Charge 2, [tex]q_2=4.39\times 10^{-6}\ C[/tex]

Distance between charges, r = 0.0209 m

1. The electric force is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}[/tex]

F = -492.95 N

2. Distance between two identical charges, [tex]r=0.0209\ m[/tex]

Electric force is given by :

[tex]F=\dfrac{kq_3^2}{r^2}[/tex]

[tex]q_3=\sqrt{\dfrac{Fr^2}{k}}[/tex]

[tex]q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}[/tex]

[tex]q_3=4.89\times 10^{-6}\ C[/tex]

Hence, this is the required solution.

Answer:

The flux of the electric field  is 677.6 Nm²/C

Explanation:

Given that,

Area = 18 cm²

Charge = 6.0 nC

We need to calculate the flux of the electric field

Using Gauss's law

[tex]\phi=\dfrac{q}{\epsilon_{0}}[/tex]

Where, q = charge

[tex]\epsilon_{0}[/tex] =permittivity of free space

Put the value into the formula

[tex]\phi=\dfrac{6.0\times10^{-9}}{8.854\times10^{-12}}[/tex]

[tex]\phi=677.6\ Nm^2/C[/tex]

Hence, The flux of the electric field  is 677.6 Nm²/C.

Answer:

F = 1080 N

Explanation:

given,

mass of the hammer = 1.8 kg

velocity of the hammer = 6 m/s

distance into board = 30 mm = 0.03 m

to calculate average resistance force = F

kinetic energy of the hammer is equal to the work done by the hammer

[tex]\dfrac{1}{2}mv^2 = Force\times displacement[/tex]

[tex]\dfrac{1}{2}mv^2 = F\times d[/tex]

[tex]\dfrac{1}{2}\times 1.8\times 6^2 = F\times 0.03[/tex]

F = 1080 N

hence, the average resistance force is equal to F = 1080 N

Answer:

The cars pass next to one another after 25.28 s.

Explanation:

When the cars pass next to one another, the position of both cars is the same relative to the center of the system of reference (marker 0 in this case). Then:

Position of car 1 = position of car 2

The position of an accelerating object moving in a straight line is given by this equation:

x = x0 +v0 t +1/2 a t²

where

x = position at time t

x0 = initial position

v0 = initial speed

t = time

a = acceleration

If the position of car 1 = position of car 2 then:

0 km + 20.0 m/s * t + 1/2 * 0.10 m/s² * t² = 1.2 km - 30.0 m/s * t + 1/2 * 0.30 m/s² * t²

Note that the acceleration of car 2 has to be positive because the car is slowing down and, in consequence, the acceleration has to be opposite to the velocity. The velocity is negative because the direction of car 2 is towards the origin of our system of reference. Let´s continue:

0 km + 20.0 m/s * t + 1/2 * 0.10 m/s² * t² = 1.2 km - 30.0 m/s * t + 1/2 * 0.30 m/s² * t²

1200 m - 50.0 m/s * t + 0.10 m/s² * t² = 0

Solving the quadratic equation:

t = 25.28 s

t = 474. 72 s We discard this value because, if we replace it in the equation of the position of car 2, we will get a position of 20762 m, which is impossible because the position of car 2 can´t be greater than 1200 m.

Then, the cars pass next to one another after 25.28 s  

Answer:1.1 eV

Explanation:

Given

Energy(E)=2.9 eV

Work function of the target=1.8 eV

We know that

[tex]Energy(E)=W_0+KE_{max}[/tex]

Where [tex]W_0=work\ function[/tex]

[tex]2.9=1.8+KE_{max}[/tex]

[tex]KE_{max}=2.9-1.8=1.1 eV[/tex]

Answer:

The answer is [tex] 3.994 \times 10^{13}\ electrons[/tex]

Explanation:

The amount of negative charge that must be removed is

[tex]\Delta = Final\ charge - initial\ Charge = 2.6 - (-3.8) = 6.4 \ \mu C = 6.4 \times 10^{-6}\ C[/tex]

and the charge of one electron is

[tex]1 e = 1.60217662\times 10^{-19} \ C[/tex]

So the amount of electrons we need to remove is

[tex]x = \frac{6.4 \times 10^{-6}}{1.60217662\times 10^{-19}} \approx 3.994 \times 10^{13}\ electrons[/tex]

Answer: 5 m/s^2

Explanation: In order to solve this question we have to use the kinematic equation given by:

Vf= Vo+a*t  where V0 is zero.

we know that it takes Vf( 30 m/s) in 6 seconds

so

a=(30 m/s)/6 s= 5 m/s^2

Answer:

(a) Your right-hand thumb must be pointed in the direction of the south pole.

Explanation:

A solenoid is a coil of insulated copper wire which behaves like a magnet when an electric current passes through it and it loses its magnetic behaviour just after the current flow stops.

Since it behaves a magnet, the magnetic poles must be determined. In order to determine the magnetic pole, we put the coil in our right hand, curl our four fingers in the direction of current flow and then the direction in which our thumb points is the north pole. This is known as the Right-hand rule.

From the above discussion, option (a) does not keep pace with the statement of right-hand rule.

Hence, option (a) is not true.

The police car takes approximately 30 seconds to reach the speeder, requires an acceleration of about 1.67 m/s², and its velocity at the moment it reaches the speeder is approximately 50 m/s.

This problem is a two-body pursuit scenario. The speeder's motion can be described by x = Ut, while the police car's motion is represented by the equation x = 1/2at². The speeder is moving at a constant speed of 120 km/h which is equal to 33.33 m/s.

(a) Time for the police car to overtake the speeder:
To find the time, we equate the two equations (since they both cover the same distance of 750m) and solve for 't'. Thus, 750m = 33.33m/s * t = 1/2 * a * t². By solving this equation, we get two values of 't', out of which the realistic answer is t = 30 seconds.

(b) Required acceleration of the police car:
Plugging the time into the equation for the police car, we get 750m = 1/2 * a * (30s)². Solving for 'a', we find the required acceleration to be approximately 1.67 m/s².

(c) Velocity of the police car at the moment it reaches the speeder:
Since the police car starts from rest and maintains a constant acceleration, the final velocity can be calculated using the equation v = at. Hence, at the moment it reaches the speeder, the police car's velocity would be approximately 50 m/s.

https://brainly.com/question/33595094

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Answer:

[tex]Density = 538 \frac{g}{m3} [/tex]

Explanation:

To get the density you need the vapor pressure for the moisture, to get this first you need to find in tables (found in internet, books, apps) the saturation vapor pressure for water at 50°C:

[tex]P_{sat} = 12350 Pa  @ 50°C[/tex]

Now, a relative humidity of 6,5% means that the actual vapor pressure is 6,5% that of the saturated air so:

[tex]P_{vapor} = 12350 Pa * 0,065 = 802,75 Pa = 0,792 atm [/tex]

According to the ideal gases formula density can be calculated as:

[tex]d = \frac{P*M}{R*T}[/tex]

Where:

Ideal gases constant [tex]R = 0,082 \frac{atm L}{mol k}[/tex]

Pressure P = 0,792 atm

Temperature T = 50°C = 323 K

molar mass M = 18 g/mol

[tex]d = 0,538 \frac{g}{L} = 538 \frac{g}{m3} [/tex]

Answer:

[tex]v(t)=2.7*e^{0.5kt}\\\\ v(s)=\frac{k}{2}*s+2.7-1.3k\\\\[/tex]

For t=2.7s and k=0.3 m/s:

[tex]v(2.7s)=1.80m/s[/tex]

For s=6m and k=0.3 m/s:

[tex]v(6m)=6.69m/s\\\\[/tex]

Explanation:

Definition of acceleration:

[tex]a=\frac{dv}{dt} =0.5kv[/tex]

we integrate in order to find v(t):

[tex]\frac{dv}{v} =-0.5kdt[/tex]

[tex]\int\limits^v_0 { \frac{dv}{v}} \, =-0.5k\int\limits^t_0 {dt} \,[/tex]

[tex]ln(v)=-0.5kt+C\\\\ v=A*e^{-0.5kt}[/tex]        A=constant

Definition of velocity:

[tex]v=\frac{ds}{dt} =A*e^{-0.5kt}[/tex]

We integrate:

[tex]v=\frac{ds}{dt} \\s=- (2/k)*A*e^{-0.5kt}+B[/tex]       B=constant

But:

[tex]v=A*e^{-0.5kt}[/tex]⇒[tex]s= -(2/k)*v+B[/tex]

[tex]v(s)=-(\frac{k}{2} )(s-B)=D-\frac{k}{2}*s[/tex]          D=other constant

Initial conditions:t = 0 are s0 = 2.6 m and v0 = 2.7 m/sec:

[tex]v(t)=A*e^{-0.5kt}\\ 2.7=Ae^{-0.5k*0}\\ 2.7=A\\[/tex]

[tex]v(s)=D-\frac{k}{2}*s\\2.7=D-\frac{k}{2}*2.6\\D=2.7+1.3k[/tex]

So:

[tex]v(t)=2.7*e^{-0.5kt}\\\\ v(s)=\frac{k}{2}*s+2.7+1.3k\\\\[/tex]

For t=2.7s and k=0.3 m/s:

[tex]v(2.7s)=2.7*e^{-0.5*0.3*2.7}=1.80m/s[/tex]

For s=6m and k=0.3 m/s:

[tex]v(6m)=\frac{0.3}{2}*6+2.7+1.3*0.3=6.69m/s\\\\[/tex]

Answer:

l= 4 mi   : width of the park

w= 1 mi  : length of the park

Explanation:

Formula to find the area of ​​the rectangle:

A= w*l       Formula(1)

Where,

A is the area of the  rectangle in mi²

w is the  width of the rectangle in mi

l is the  width of the rectangle in mi

Known data

A =  4 mi²

l = (w+3)mi    Equation (1)

Problem development

We replace the data in the formula (1)

A= w*l  

4 = w* (w+3)

4= w²+3w

w²+3w-4= 0

We factor the equation:

We look for two numbers whose sum is 3 and whose multiplication is -4

(w-1)(w+4) = 0 Equation (2)

The values ​​of w for which the equation (2) is zero are:

w = 1 and w = -4

We take the positive value w = 1 because w is a dimension and cannot be negative.

w  = 1 mi  :width of the park

We replace w  = 1 mi  in the equation (1) to calculate the length of the park:

l=  (w+3) mi

l= ( 1+3) mi

l= 4 mi

Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

[tex]F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}[/tex]

Where:

G= 6.674×10^−11 N · (m/kg)2

k =  8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

[tex]\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}[/tex]   --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is  5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

[tex]\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}[/tex]

Therefore

[tex]\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}[/tex]

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field

Answer:

[tex]q=1.18*10^{-6}C}[/tex]

Explanation:

The potential V due to a charge q,  at a distance r, is:

[tex]V=k\frac{q}{r}[/tex]

k=8.99×109 N·m^2/C^2      :Coulomb constant

We solve to find q:

[tex]q=\frac{V*r}{k}=\frac{6.25*10^{2}*17}{8.99*10^{9}}=1.18*10^{-6}C[/tex]

Answer:

a) 0.2399 mi³

b) 440.8 × 10³ Pounds

Explanation:

Given:

Volume of cumulus cloud, V = 1 km³

Liquid water content = 0.2 g/m³

Now,

a) 1 km = [tex]\frac{\textup{1 miles}}{\textup{1.6093}}[/tex]

thus,

1 km³ = [tex](\frac{\textup{1 miles}}{\textup{1.6093}})^3[/tex]

1 km³ =  0.2399 mi³

Hence, volume of cloud in cubic miles is 0.2399 mi³

b)

Liquid water content = 0.2 g/m³

Now,

1 Km = 1000 m

thus,

1 km³ = 1000³ m³

Therefore,

Liquid water content in 1 Km³ of cloud = 0.2 g/m³ × 1000³ m³

= 200 × 10⁶ gram

or

= 200 × 10³ Kg

also,

1 kilogram =  2.204 pounds

Therefore,

200 × 10³ Kg = 200 × 10³ × 2.204 pounds = 440.8 × 10³ Pounds

The volume of a cumulus cloud occupying one cubic kilometer is equal to 0.386102 cubic miles. The water inside this cloud, with a liquid water content of 0.2 g/m³, weighs approximately 441 pounds.

The subject of this problem is to find the volume of a cloud in a different unit of measure and the weight of the water within it. Volume conversion from cubic kilometers to cubic miles and weight calculation of water in pounds are required.

To find the volume of the cloud in cubic miles, we use the conversion factor that 1 cubic kilometer is equal to about 0.386102 cubic miles. Hence, the volume of 1 cubic kilometer in cubic miles is:

1 cubic kilometer * 0.386102 cubic miles/cubic kilometer = 0.386102 cubic miles.

Next, we calculate the weight of the water in pounds, considering that the liquid water content is 0.2 grams per cubic meter and there are 1,000,000 cubic meters in a cubic kilometer:

0.2 g/m³ * 1,000,000 m³/km³ = 200,000 grams of water in the cloud.

Now, convert grams to pounds using the conversion factor 453.59237 grams per pound:

200,000 grams * (1 pound / 453.59237 grams) = 441 pounds.

Therefore, the water weight in the cloud is 441 pounds.

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

[tex]F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}[/tex].

Now, making the substitutions

[tex]d \ = \ 2.00 * 10 ^{-15} \ m[/tex],

[tex]q_1 = q_2 = 1.60 * 10 ^ {-19} \ C[/tex],

[tex]\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2}[/tex],

we can find:

[tex]F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}[/tex].

[tex]F \ = 57.536 N[/tex].

Not so big for everyday life, but enormous for subatomic particles.

The magnitude of the electric force between two protons separated by 2.00 × 10⁻¹⁵ m is about 57.6 Newton

[tex]\texttt{ }[/tex]

Electric charge consists of two types i.e. positively electric charge and negatively electric charge.

[tex]\texttt{ }[/tex]

There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :

[tex]\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }[/tex]

F = electric force (N)

k = electric constant (N m² / C²)

q = electric charge (C)

r = distance between charges (m)

The value of k in a vacuum = 9 x 10⁹ (N m² / C²)

Let's tackle the problem now !

[tex]\texttt{ }[/tex]

Given:

distance between protons = d = 2 × 10⁻¹⁵ m

charge of proton = q = 1.6 × 10⁻¹⁹ C

Unknown:

electric force = F = ?

Solution:

[tex]F = k \frac{q_1 q_2}{(d)^2}[/tex]

[tex]F = k \frac{q^2}{(d)^2}[/tex]

[tex]F = 9 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{(2 \times 10^{-15})^2}[/tex]

[tex]F = 57.6 \texttt{ Newton}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Static Electricity

Answer:

Ax  =7 cm

Explanation:

Analysis:

Let A be the vector of rectangular components Ax and Ay:

Where:

Ay=7 cm  : The The y-component of  vector A-component of  vector A

α =45       : Angle of A with the positive x-axis

Ax           : The x-component of  vector A

Because Ax and Ay are the components of a right triangle we apply the following formula:

[tex]tan\alpha =\frac{A_{y} }{A_{x} }[/tex]

[tex]tan45=\frac{7}{A_{x} }[/tex]

[tex]A_{x} =\frac{7}{tan45}[/tex]

Ax  =7 cm